how do the graphs of y= 1/x and y= 5/x+6 compare?
Go on:
wolframalpha.c o m
When page be open in rectangle type:
plot y= 1/x
and clck option =
you will see graph
Then type:
plot y= 5/(x+6)
and clck option =
___________________________________
Remark:
You must type plot y= 5/(x+6) with brackets
___________________________________
each has a vertical and a horizontal asymptote.
you can supply the where ...
could u explain it more
To compare the graphs of y = 1/x and y = 5/(x + 6), we can start by analyzing each equation separately and plotting their corresponding graphs.
For the equation y = 1/x, we need to find several points to plot the graph. Let's select some values for x and then calculate the corresponding y-values using the equation:
When x = -3, y = 1/(-3) = -1/3
When x = -2, y = 1/(-2) = -1/2
When x = -1, y = 1/(-1) = -1
When x = 1, y = 1/1 = 1
When x = 2, y = 1/2 = 1/2
When x = 3, y = 1/3 = 1/3
Plotting these points on a graph, we can see that y = 1/x is a hyperbola with the x-axis and y-axis acting as asymptotes. The graph lies in Quadrant I and Quadrant III.
For the equation y = 5/(x + 6), we will follow the same process to find some points:
When x = -9, y = 5/(-9 +6) = -5/3
When x = -6, y = 5/(-6 +6) = undefined (division by zero is not possible)
When x = -3, y = 5/(-3 +6) = 5/3
When x = 0, y = 5/(0 +6) = 5/6
When x = 3, y = 5/(3 +6) = 5/9
When x = 6, y = 5/(6 +6) = 5/12
Now we can plot these points. The graph of y = 5/(x + 6) is a hyperbola as well, but with a vertical shift of 6 units to the left compared to y = 1/x. It also has vertical and horizontal asymptotes defined by the equations x = -6 and y = 0, respectively.
By comparing the two graphs, you can see that they have similar shapes but are shifted horizontally. The graph of y = 5/(x + 6) is shifted 6 units to the left compared to y = 1/x.